Answer:
0.50 M
Explanation:
Given data
- Mass of sodium sulfate (solute): 7.1 g
- Volume of solution: 100 mL
Step 1: Calculate the moles of the solute
The molar mass of sodium sulfate is 142.04 g/mol. The moles corresponding to 7.1 grams of sodium sulfate are:

Step 2: Convert the volume of solution to liters
We will use the relation 1 L = 1000 mL.

Step 3: Calculate the molarity of the solution

Answer:
D. I < III < II
Explanation:
- The osmotic pressure (π) is given by the relation:
<em>π = iMRT.</em>
where, π is the osmotic pressure.
i is van 't Hoff factor.
M is the molarity of the solution.
R is the general gas constant.
T is the temperature.
<em>M, R and T are constant for all solutions.</em>
So, the osmotic pressure depends on the van 't Hoff factor.
- The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass.
- For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.
- For most ionic compounds dissolved in water, the van 't Hoff factor is equal to the number of discrete ions in a formula unit of the substance.
For C₂H₆O₂ (non-electrolyte solute): i = 1.
For MgCl₂: i = 3.
It dissociates to give (Mg²⁺ + 2Cl⁻).
For NaCl: i = 2.
It dissociates to give (Na⁺ + Cl⁻).
So, the solute that has the highest osmotic pressure is II. 0.15 M MgCl₂, then III. 0.15 M NaCl, then I. 0.15 M C₂H₆O₂.
<em>D. I < III < II.</em>
<em></em>
Answer: a)
and pH = 1.04
b)
and 
c)
and ![[OH^-]=1.41\times 10^{-4}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.41%5Ctimes%2010%5E%7B-4%7D)
Explanation:
pH or pOH is the measure of acidity or alkalinity of a solution.
pH is calculated by taking negative logarithm of hydrogen ion concentration.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)
![pOH=-\log [OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%5BOH%5E-%5D)

a) ![[H^+]=0.090M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.090M)
![pH=-\log [0.090]=1.04](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5B0.090%5D%3D1.04)

![12.96=-log[OH^-]](https://tex.z-dn.net/?f=12.96%3D-log%5BOH%5E-%5D)
![[OH^-]=1.09\times 10^{-13}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.09%5Ctimes%2010%5E%7B-13%7D)
b) ![[OH^-]=0.00098M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.00098M)
![pOH=-\log [0.00098]=3.01](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%5B0.00098%5D%3D3.01)

![10.99=-log[H^+]](https://tex.z-dn.net/?f=10.99%3D-log%5BH%5E%2B%5D)
![[H^+]=1.02\times 10^{-11}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D1.02%5Ctimes%2010%5E%7B-11%7D)
c) 
![10.15=-\log [H^+]](https://tex.z-dn.net/?f=10.15%3D-%5Clog%20%5BH%5E%2B%5D)
![[H^+]=7.08\times 10^{-11}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D7.08%5Ctimes%2010%5E%7B-11%7D)

![3.85=-log[OH^-]](https://tex.z-dn.net/?f=3.85%3D-log%5BOH%5E-%5D)
![[OH^-]=1.41\times 10^{-4}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.41%5Ctimes%2010%5E%7B-4%7D)
Two electron pairs is the answer
Answer:
See Explanation
Explanation:
8. 3, 1, -1, +1/2 => 3pₓ¹ => Aluminum
9. 4, 2, +1, +1/2 => 4d₁¹ => Chromium
10. 6, 1, 0, -1/2 => 6p₀² => Argon
11. 4, 3, +3, -1/2 => 4f₊₃² => Lutetium
12. 2, 1, +1, -1/2 => 2p₊₁² => Neon